CONDITIONING AND EVIDENCE Company A has just developed a diagnostic test for a certain disease. The disease a?icts 1% of the population. As de?ned in Example 2.3.9, the sensitivity of the test is the...

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CONDITIONING AND EVIDENCE
Company A has just developed a diagnostic test for a certain disease. The disease a?icts 1% of the population. As de?ned in Example 2.3.9, the sensitivity of the test is the probability of someone testing positive, given that they have the disease, and the speci?city of the test is the probability that of someone testing negative, given that they don’t have the disease. Assume that, as in Example 2.3.9, the sensitivity and speci?city are both 0.95. Company B, which is a rival of Company A, o?ers a competing test for the disease. Company B claims that their test is faster and less expensive to perform than Company A’s test, is less painful (Company A’s test requires an incision), and
2
yet has a higher overall success rate, where overall success rate is de?ned as the probability that a random person gets diagnosed correctly.

(a) It turns out that Company B’s test can be described and performed very simply: no matter who the patient is, diagnose that they do not have the disease. Check whether Company B’s claim about overall success rates is true.

(b) Explain why Company A’s test may still be useful.

(c) Company A wants to develop a new test such that the overall success rate is higher than that of Company B’s test. If the sensitivity and speci?city are equal, how high does the sensitivity have to be to achieve their goal? If (amazingly) they can get the sensitivity equal to 1, how high does the speci?city have to be to achieve their goal? If (amazingly) they can get the speci?city equal to 1, how high does the sensitivity have to be to achieve their goal?
Answered Same DayDec 26, 2021

Answer To: CONDITIONING AND EVIDENCE Company A has just developed a diagnostic test for a certain disease. The...

David answered on Dec 26 2021
126 Votes
Solution
(a)
Here, we see that the probability of success for company A is found to be 0.95, so
the overall
probability must be 0.95
P(success) = P(A|E)P(E) + P(A
C
|E
C
)P(E
C
) = 0.95 * 0.01 + 0.95 * 0.99 = 0.95
The probability for success for company B is 0.99
P(success) = P(B|E)P(E) +...
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